Honda Activa accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.
Answers
Answered by
17
Answer:
0.71 m/s²
Explained:
Honda Activa was initially at rest, where its velocity was 0.
Later on, when it had covered 35.4 m, velocity becomed 7.10.
From this,
Initial velocity = 0
Final velocity = 7.10 m/s
Displacement = 35.4 m
Using equations of motion:
• v^2 = u^2 + 2aS
= > (7.10)² = 0² + 2a(35.4)
= > 50.41 = 70.8a
= > 50.41/70.8 = a
= > 0.71 m/s² ≈ a
Answered by
20
Answer:
0.71 m/s²
Explanation:
Given; initial velocity is 0 m/s, final velocity is 7.10 m/s and distance covered is 35.4 m.
To find: Acceleration of the bike.
Using the third equation of motion,
v² - u² = 2as
Substitute the values,
→ (7.10)² - (0)² = 2a(35.4)
→ 50.41 = 70.8a
Divide by 70.8 on both sides,
→ 50.41/70.8 = 70.8a/70.8
On solving we get,
→ 0.71 m/s² = a
Hence, the acceleration of the bike is 0.71 m/s².
Similar questions